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3 years ago

A dilation with a scale factor of 2 is applied to the image below. What is the length of the new image?

Mathematics

1 answer:

STatiana [176]3 years ago

8 0

A dilation of factor k multiplies all lengths by k.

So, in your case, all lengths double. The new length is 4.

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Express the ratio as a fraction in simplest form. 15:45 2/3 3/9 1/3 5/16

ankoles [38]

Answer:

3/9

Step-by-step explanation:

15/45

divide both sides by a factor of each number; I'll be using 5.

3/9

It didn't have to be 5, I could have used 3, but in that case It wouldn't have been it's simplest form and I would have to divide again by another factor.

3 0

2 years ago

What is the equation in slope-intercept form of the line that has a slope of -2/3, and passes through the point (6, 2)?

Jet001 [13]

Answer:

Y=-2/3x+6

Step-by-step explanation:

Y=-2/3x+b

2=-2/3•6+b

2=-4+b

6=b

3 0

3 years ago

Solve -r/4 < 8 A.r<-32 B. r. >-32 C.r < -2 D.r>-2

Gnom [1K]

$\frac{-r}{4}$

$Ans.\ B$

4 0

3 years ago

keisha bought 1/6 tank of gas in the morning on her way to work. she added 3/4 tank on her way home from work. what fraction of

larisa [96]

1/6 x 2 = 2/12
3/4 x 3 = 9/12
2/12 + 9/12 = 11/12

Altogether, Keisha bought 11/12 of a tank of gas.

6 0

3 years ago

A sample of a radioactive substance decayed to 97% of its original amount after a year. (Round your answers to two decimal place

Andrej [43]

Answer:

a) The half life of the substance is 22.76 years.

b) 5.34 years for the sample to decay to 85% of its original amount

Step-by-step explanation:

The amount of the radioactive substance after t years is modeled by the following equation:

$P(t) = P(0)(1-r)^{t}$

In which P(0) is the initial amount and r is the decay rate.

A sample of a radioactive substance decayed to 97% of its original amount after a year.

This means that:

$P(1) = 0.97P(0)$

Then

$P(t) = P(0)(1-r)^{t}$

$0.97P(0) = P(0)(1-r)^{0}$

$1 - r = 0.97$

$P(t) = P(0)(0.97t)^{t}$

(a) What is the half-life of the substance?

This is t for which P(t) = 0.5P(0). So

$P(t) = P(0)(0.97t)^{t}$

$0.5P(0) = P(0)(0.97t)^{t}$

$(0.97)^{t} = 0.5$

$\log{(0.97)^{t}} = \log{0.5}$

$t\log{0.97} = \log{0.5}$

$t = \frac{\log{0.5}}{\log{0.97}}$

$t = 22.76$

The half life of the substance is 22.76 years.

(b) How long would it take the sample to decay to 85% of its original amount?

This is t for which P(t) = 0.85P(0). So

$P(t) = P(0)(0.97t)^{t}$

$0.85P(0) = P(0)(0.97t)^{t}$

$(0.97)^{t} = 0.85$

$\log{(0.97)^{t}} = \log{0.85}$

$t\log{0.97} = \log{0.85}$

$t = \frac{\log{0.85}}{\log{0.97}}$

$t = 5.34$

5.34 years for the sample to decay to 85% of its original amount

8 0

3 years ago